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Let $(W_t)$ be a Brownian motion with respect to a filtration $(\mathcal{F}_t)$. For all $t \geq 0 $ set $$X_t = \int_0^t W_s^2 \mathrm{d} W_s,\qquad Y_t = W_t^7.$$ Find the covariance process $\langle X,Y \rangle_t$ for all $t \geq 0$.

My idea is (since $X_t $ and $Y_t$ are cont. local martingales) to use that

$$\langle X,Y \rangle_t = \langle (W_s^2 \cdot W_s ), (1 \cdot W_t^7) \rangle_t = \int_0^t W_s^2 \mathrm{d} \langle W,W^7 \rangle_s $$

But I don´t know how to calculate the integral above. Should I find $\langle W,W^7 \rangle_t,$ where I use that $\langle W,W^7 \rangle_t,$ makes $W_tW_t^7 - \langle W,W^7 \rangle_t, = W_t^8 - \langle W,W^7 \rangle_t$ a continuous local martingale?

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  • $\begingroup$ Got something from the answer below? $\endgroup$
    – Did
    Commented Oct 17, 2014 at 10:11

1 Answer 1

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Why not going for the simplest? You are given that $\mathrm dX_t=\color{red}{W^2_t}\cdot\mathrm dW_t$ and you should be able to compute $$\mathrm dY_t=\color{red}{7W^6_t}\cdot\mathrm dW_t+21W^5_t\cdot\mathrm dt,$$ hence $$\mathrm d\langle X,Y\rangle_t=\color{red}{W^2_t}\cdot\color{red}{7W^6_t}\cdot\mathrm d\langle W,W\rangle_t=7W^8_t\cdot\mathrm dt,$$ or, equivalently, $$\langle X,Y\rangle_t=7\int_0^tW^8_s\mathrm ds.$$

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  • $\begingroup$ But why does $21 W_t^5 \cdot dt$ don´t appear in $d \langle X,Y \rangle$ ? $\endgroup$
    – John
    Commented Oct 15, 2014 at 13:37
  • $\begingroup$ Because, as your textbook indicates, if $dX_t=\sigma_t dW_t+a_tdt$ and $dY_t=\tau_t dW_t+b_tdt$ then $d\langle X,Y\rangle_t=\sigma_t\tau_tdt$. $\endgroup$
    – Did
    Commented Oct 15, 2014 at 13:39

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